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How do communication systems convert, transmit and receive information, and how do engineering principles inform their design?

Investigate the elements of a communications system (information source, transmitter, channel, receiver, destination), analyse analogue and digital signals, and apply principles such as modulation, bandwidth, signal-to-noise ratio, attenuation and multiplexing

A focused HSC Engineering Studies Telecommunications Engineering answer on communication systems fundamentals. Covers the 5-element system model, analogue vs digital signals, modulation (AM/FM/PM/digital), bandwidth, S/N, attenuation, multiplexing (TDM/FDM), and engineering implications.

Generated by Claude Opus 4.89 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA Telecommunications Engineering Module (HSC, one of four compulsory modules per the 2011 Engineering Studies syllabus alongside Civil Structures, Personal and Public Transport, and Aeronautical Engineering) starts with the fundamentals of communication systems. You need the standard 5-element model, the analogue / digital distinction, and the core engineering parameters (modulation, bandwidth, S/N, attenuation, multiplexing) with their practical engineering implications.

The answer

The 5-element communication system model

Every communication system can be decomposed into five elements:

  1. Information source. What generates the message (a speaker's voice; a camera; a sensor; a computer).
  2. Transmitter. Converts the source's signal into a form suitable for transmission. Performs modulation, amplification, encoding.
  3. Channel. The physical medium that carries the signal (copper wire, coaxial cable, optical fibre, free space for radio).
  4. Receiver. Inverts what the transmitter did. Performs demodulation, amplification, decoding, error correction.
  5. Destination. Where the received message ends up (a speaker; a screen; a control system).

Real systems often add repeaters (amplifying and re-transmitting signals along long channels) and noise (entering at any stage but most consequentially in the channel).

The five-element communication system model A schematic block diagram showing five boxes in a row connected by arrows: information source, transmitter, channel, receiver, destination. A dashed box labelled noise points into the channel box from below, and a smaller box labelled repeater sits above the channel box with arrows both into and out of it, showing it amplifies and re-transmits the signal partway along the channel. Information source Transmitter Channel wire, fibre, or air Receiver Destination Repeater Noise Transmitter and receiver mirror each other; the channel is where noise and attenuation dominate.

Analogue vs digital signals

Analogue signals vary continuously in amplitude and time. Examples: voice on traditional telephone lines; FM radio broadcast.

Digital signals take discrete values (typically two, 0 and 1) sampled at discrete times. Examples: text messages, internet traffic, modern voice calls (after analogue-to-digital conversion).

Advantages of digital:

  • Noise immunity. Digital signals can be regenerated cleanly because the receiver only needs to distinguish two states. Analogue signals degrade cumulatively.
  • Compression. Digital data can be compressed; analogue cannot.
  • Encryption. Digital data can be encrypted; analogue cannot easily.
  • Multiplexing. Multiple digital streams can share a channel via time-division multiplexing more easily than analogue.
  • Error correction. Digital data can carry redundancy bits that let receivers detect and correct errors.

Advantages of analogue (where retained):

  • Bandwidth efficiency for some specific applications (FM radio remains analogue).
  • Simplicity of the basic transmitter/receiver for some legacy systems.
  • Continuous fidelity in principle (no quantisation error), though noise eventually dominates.

The general direction of telecommunications engineering since the 1980s has been progressive digitisation: voice telephony, broadcast TV, radio, photography, and data have all moved from analogue to digital.

Modulation

Modulation is the process of impressing the information signal onto a higher-frequency carrier wave so it can be transmitted efficiently. Demodulation at the receiver recovers the original signal.

Analogue modulation:

  • AM (Amplitude Modulation). The amplitude of the carrier varies with the information signal. Used in AM radio (530-1700 kHz). Vulnerable to amplitude noise; lower fidelity than FM.
  • FM (Frequency Modulation). The frequency of the carrier varies with the information signal. Used in FM radio (88-108 MHz), traditional TV audio. Less vulnerable to amplitude noise; higher fidelity.
  • PM (Phase Modulation). The phase of the carrier varies with the information signal. Used in some digital systems as a base for PSK.

Digital modulation:

  • ASK (Amplitude Shift Keying). Bits represented by different amplitudes.
  • FSK (Frequency Shift Keying). Bits represented by different frequencies. Used in early modems.
  • PSK (Phase Shift Keying). Bits represented by different phases. Used in WiFi, 4G, satellite links.
  • QAM (Quadrature Amplitude Modulation). Combines amplitude and phase to carry multiple bits per symbol. Used in cable modems, digital TV, 5G.

Bandwidth

Bandwidth is the range of frequencies a signal occupies, measured in Hz. Higher-bandwidth channels can carry more information per second.

  • AM radio channel: ~10 kHz bandwidth per station.
  • FM radio channel: ~200 kHz bandwidth per station.
  • Voice telephone channel: ~3.4 kHz bandwidth.
  • Fibre optic channel: gigahertz-to-terahertz bandwidth potential.

Engineering trade-off: higher bandwidth means more data capacity but also more noise picked up and more channel occupied.

Signal-to-noise ratio (S/N)

S/N is the ratio of signal power to noise power, usually expressed in decibels (dB):

S/N (dB) = 10 log_10 (Psignal / Pnoise)

Higher S/N = cleaner reception. Below a threshold S/N, demodulation fails. Engineers design systems with margin above the threshold.

Sources of noise in a communication system: thermal noise (electronic components), interference (other transmissions on nearby frequencies), atmospheric noise (lightning, solar activity), shot noise (in semiconductor devices).

Attenuation

Attenuation is the loss of signal power as it travels through a channel, measured in dB per unit distance (e.g. dB/km for fibre optic; dB per 100m for copper).

  • Copper twisted pair: ~6 dB/km at voice frequencies.
  • Coaxial cable: ~10-30 dB/km depending on frequency.
  • Single-mode fibre optic: ~0.2 dB/km (at 1550 nm; the reason fibre dominates long-distance).
  • Free space radio: varies with frequency, atmosphere, and distance (inverse-square losses plus atmospheric absorption).

Repeaters compensate for attenuation by amplifying and re-transmitting along the link.

Multiplexing

Multiplexing allows multiple signals to share a single channel.

  • Frequency Division Multiplexing (FDM). Each signal occupies a different frequency band. Examples: AM radio broadcast (many stations on different frequencies sharing the broadcast spectrum); cable TV.
  • Time Division Multiplexing (TDM). Each signal occupies a different time slot. Examples: traditional digital telephony; some satellite systems.
  • Code Division Multiplexing (CDM). Signals share the same frequency and time but are distinguished by different codes. Used in some mobile standards.
  • Wavelength Division Multiplexing (WDM). Optical fibre version of FDM; different colours of light carry different signals. Enables single-fibre data rates beyond 1 Tbps.

Examples in context

Example 1. Australian NBN as a multiplexed system. Australia's National Broadband Network combines technologies based on distance and existing infrastructure: fibre to the premises (FTTP) where economical; fibre to the node (FTTN) with copper to the home; hybrid fibre-coaxial (HFC); fixed wireless; and satellite for remote areas. Each leg uses different modulation and multiplexing. The NBN illustrates engineering trade-offs at national scale: rural areas accept lower bandwidth (satellite) because trenching fibre across continental distances is uneconomic.

Example 2. Mobile network evolution. GSM (2G, 1990s) used TDMA at ~9.6 kbps voice and basic text. 3G added CDMA and reached ~2 Mbps. 4G introduced OFDM and reached ~100 Mbps. 5G uses massive MIMO and millimetre-wave bands to reach ~1 Gbps. Each generation shifted modulation and multiplexing to extract more bandwidth from finite spectrum. The progression illustrates how engineering parameters (modulation order, bandwidth, multiplexing scheme) drive performance.

Try this

Q1. Identify the five elements of a communication system. [2 marks]

  • Cue. Information source; transmitter; channel; receiver; destination.

Q2. Distinguish between AM and FM modulation, including one advantage of each. [4 marks]

  • Cue. AM = amplitude varies. Advantage: simpler / less bandwidth per station. FM = frequency varies. Advantage: less vulnerable to amplitude noise / higher fidelity.

Q3. Justify the choice of optical fibre over copper twisted pair for a long-distance telecommunications backbone. [6 marks]

  • Cue. Attenuation (0.2 dB/km vs ~50 dB/km broadband copper); bandwidth (gigahertz-terahertz vs MHz); resistance to electromagnetic interference; security (harder to tap); future-proofing for data growth. Acknowledge trade-off: higher initial deployment cost.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 HSC2 marksOutline how GPS satellites determine a position on the planet.
Show worked answer →

A 2 mark outline needs the idea of timing signals from several satellites and calculating from them.

Position is determined by trilateration: the GPS receiver measures the time it takes signals to travel from several satellites to the receiver, then uses the differences in these travel times (and the known satellite positions) to calculate its location. At least three satellites are required, the satellites only transmit the signal, and the calculation is performed by the GPS device itself.

Markers reward mentioning multiple satellites, signal travel time, and that the receiver performs the calculation.

2023 HSC1 marksWhat is the primary benefit of placing a satellite in a low earth orbit rather than a higher earth orbit? A. Longer lifespan B. Less atmospheric drag C. Faster data transmission D. Cover larger areas on Earth
Show worked answer →

The correct answer is C. A satellite in a low earth orbit is much closer to the ground, so signals travel a shorter distance, giving lower latency and faster data transmission.

A and B are actually disadvantages of low earth orbit (shorter lifespan and more atmospheric drag because the satellite is lower in the atmosphere). D is a benefit of a higher orbit, not a low one, because a higher satellite can see and cover a larger area of the Earth.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksIdentify the five elements of a communication system and give one real example for each, using a smartphone voice call as the scenario.
Show worked solution →
  1. Information source: the caller's voice.
  2. Transmitter: the phone's microphone and radio circuitry, which converts and modulates the voice signal.
  3. Channel: the air (radio channel) between the phone and the mobile tower, and onward through the network.
  4. Receiver: the recipient's phone radio circuitry, which demodulates the signal.
  5. Destination: the recipient's speaker (or ear).

Marking criteria: 1 mark for correctly naming all five elements in order, 1 mark for a plausible smartphone-specific example for at least three elements, 1 mark for correctly matching source/destination as the human ends and transmitter/receiver as the phone hardware.

foundation3 marksCalculate the signal-to-noise ratio in decibels for a receiver with signal power 4.0 mW and noise power 0.02 mW, and state whether this link is likely to demodulate reliably.
Show worked solution →

Step 1: apply the formula.

S/N(dB)=10log⁡10(4.00.02)=10log⁡10(200)=10×2.301=23.0 dBS/N (dB) = 10 \log_{10}\left(\frac{4.0}{0.02}\right) = 10 \log_{10}(200) = 10 \times 2.301 = 23.0\ \text{dB}

Step 2: interpret. 23 dB is a healthy positive margin (well above 0 dB, where signal and noise power would be equal), so this link is very likely to demodulate reliably, provided the receiver's required threshold S/N for its modulation scheme is below about 23 dB.

Marking criteria: 1 mark for correctly setting up the ratio inside the log, 1 mark for the correct numeric answer to 1 decimal place, 1 mark for a reasoned interpretation linking the value to reliable demodulation.

core5 marksThe table below gives typical attenuation for three transmission media. A link must carry a signal 25 km with no more than 30 dB total loss before it needs a repeater. Medium | Attenuation (dB/km) --- | --- Copper twisted pair | 6 Coaxial cable | 15 Single-mode fibre | 0.2. (a) Calculate the total attenuation over 25 km for each medium. (b) State which medium(s) meet the 30 dB limit without a repeater.
Show worked solution →

(a) Total attenuation = (dB/km) x distance (km).

Copper=6×25=150 dB\text{Copper} = 6 \times 25 = 150\ \text{dB}

Coaxial=15×25=375 dB\text{Coaxial} = 15 \times 25 = 375\ \text{dB}

Fibre=0.2×25=5 dB\text{Fibre} = 0.2 \times 25 = 5\ \text{dB}

(b) Which meet the 30 dB limit. Only single-mode fibre (5 dB) is within the 30 dB budget; copper (150 dB) and coaxial (375 dB) both far exceed it and would require multiple repeaters along the 25 km run.

Marking criteria: 1 mark each for the three correct total-attenuation calculations, 1 mark for correctly identifying only fibre meets the limit, 1 mark for stating that copper and coaxial would require repeaters (with a comparative reason such as fibre's much lower dB/km).

core4 marksExplain why frequency division multiplexing (FDM) is well suited to AM radio broadcasting, and describe one drawback of FDM compared with time division multiplexing (TDM) for digital data.
Show worked solution →

FDM suits AM radio broadcasting because each station is permanently and simultaneously "on air": giving every station its own fixed frequency band (its allocated channel, e.g. spaced across the 530 to 1700 kHz AM band) lets receivers tune to any station at any time without needing to coordinate time slots, and analogue radio hardware (simple tuned circuits) can select a frequency band cheaply.

One drawback of FDM compared with TDM for digital data is spectral inefficiency when traffic is bursty: FDM permanently reserves a frequency band for each user or signal whether or not it is actively transmitting, whereas TDM can allocate time slots dynamically, letting idle users free up capacity for active ones, which better matches how digital data traffic (e.g. internet packets) tends to arrive in bursts rather than continuously.

Marking criteria: 1 mark for linking FDM to always-on, simultaneous analogue broadcast, 1 mark for a correct reason (no time coordination needed / simple tuned receivers), 1 mark for identifying an FDM drawback (spectral inefficiency with bursty traffic), 1 mark for a clear TDM-based contrast.

core4 marksAn engineer must choose between PSK and QAM for a new digital microwave link with a good, stable signal-to-noise ratio and a requirement for high data rate in limited bandwidth. Justify the more appropriate choice.
Show worked solution →

QAM is the more appropriate choice. QAM varies both amplitude and phase, so each transmitted symbol can represent more bits than a PSK symbol that only varies phase (for example, 16-QAM carries 4 bits per symbol versus 1 to 2 bits per symbol for simple PSK schemes), which directly increases data rate for the same bandwidth and symbol rate.

The trade-off is that QAM's amplitude variations make it more sensitive to noise than PSK, because a noisy receiver can more easily misread which amplitude/phase combination was sent. Since the link described has a good, stable SNR, this sensitivity is not a limiting concern, so QAM's higher spectral efficiency can be exploited without an unacceptable increase in error rate.

Marking criteria: 1 mark for correctly choosing QAM, 1 mark for explaining the higher bits-per-symbol mechanism, 1 mark for correctly noting QAM's greater noise sensitivity as the trade-off, 1 mark for linking the choice explicitly to the stable, good SNR condition given in the stem.

exam6 marksEvaluate the engineering trade-offs in choosing between fully digital and fully analogue transmission for a new regional AM/FM-equivalent radio broadcast service, considering signal quality, cost, and spectrum efficiency.
Show worked solution →

This is a 6-mark EVALUATE: markers reward a balanced weighing of both sides culminating in a supported judgement, not a one-sided description.

Case for digital broadcast
Digital broadcast signals can be regenerated cleanly at each stage because the receiver only needs to distinguish discrete symbol states, so signal quality does not progressively degrade with distance or minor interference the way analogue does; digital audio can also be compressed, letting several stations' worth of programming share spectrum that would otherwise carry only one analogue station, improving spectrum efficiency; and digital streams can carry metadata (song titles, traffic alerts) that analogue cannot.
Case for analogue broadcast
Analogue transmitters and, critically, the enormous installed base of analogue receivers are far cheaper, since no new receiver hardware is required in every listener's home or car; analogue FM in particular already achieves good perceived audio quality at modest bandwidth (~200 kHz) with mature, low-cost circuit designs; and a regional service may have too small an audience to justify the cost of digital transmission infrastructure and a broadcast standard changeover.
Judgement
For a new regional service, the deciding factor is the installed receiver base and audience size rather than pure signal quality: if most local receivers are legacy analogue radios (common in regional areas), the cost of stranding that audience or requiring a receiver upgrade generally outweighs digital's spectrum-efficiency and quality gains, so analogue (or a simulcast/parallel-run) is the more defensible short-to-medium-term engineering choice, with a planned digital transition once receiver penetration and network economics justify it.

Marker's note: top-band answers explicitly weigh multiple named factors on both sides (quality, cost, spectrum efficiency, receiver base) and finish with an explicit, reasoned recommendation, rather than only listing generic pros and cons of "digital" and "analogue" in isolation.

Openly AI-built from the official NESA syllabus, past papers and marking guides. Not individually reviewed; always check NESAand your school's materials for current rules.

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